(a)(i) Ans: 106.0°
Using the cosine rule in \(\triangle BCD\): \(\cos(\angle CBD) = \frac{BC^2 + BD^2 – CD^2}{2 \times BC \times BD}\). Substituting the given values, \(\angle CBD = \cos^{-1}(-0.275) \approx 106.0^\circ\).
(a)(ii) Ans: 292.0°
The bearing of D from B is \(038^\circ\). Since \(\angle CBD = 106.0^\circ\), the bearing of C from B is \(038^\circ + 180^\circ + 74.0^\circ = 292.0^\circ\).
(a)(iii) Ans: 310.0°
A is due east of B, so the bearing of D from A is \(360^\circ – \angle DAB\). Using the sine rule in \(\triangle ABD\), \(\angle DAB \approx 50.0^\circ\), giving a bearing of \(310.0^\circ\).
(b)(i) Ans: 15\,500\,\text{m}^2\)
Using the area formula for \(\triangle BCD\): \(\text{Area} = \frac{1}{2} \times BC \times BD \times \sin(\angle CBD)\). Substituting the values, \(\text{Area} \approx 15\,500\,\text{m}^2\).
(b)(ii) Ans: \$55\,400
Convert the area to hectares: \(15\,500\,\text{m}^2 = 1.55\,\text{ha}\). Multiply by the cost per hectare: \(1.55 \times 35\,750 \approx 55\,400\) (nearest \$100).
